Integrand size = 8, antiderivative size = 74 \[ \int e^{\text {sech}^{-1}(a x)} x \, dx=\frac {x}{a}+\frac {1}{2} \sqrt {-1+\frac {1}{a x}} \sqrt {1+\frac {1}{a x}} x^2-\frac {\arctan \left (\sqrt {-1+\frac {1}{a x}} \sqrt {1+\frac {1}{a x}}\right )}{2 a^2} \] Output:
x/a+1/2*(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2)*x^2-1/2*arctan((-1+1/a/x)^(1/2)*( 1+1/a/x)^(1/2))/a^2
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.01 \[ \int e^{\text {sech}^{-1}(a x)} x \, dx=\frac {2 a x+a x \sqrt {\frac {1-a x}{1+a x}} (1+a x)+i \log \left (-2 i a x+2 \sqrt {\frac {1-a x}{1+a x}} (1+a x)\right )}{2 a^2} \] Input:
Integrate[E^ArcSech[a*x]*x,x]
Output:
(2*a*x + a*x*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x) + I*Log[(-2*I)*a*x + 2*Sq rt[(1 - a*x)/(1 + a*x)]*(1 + a*x)])/(2*a^2)
Time = 0.34 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.72, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6889, 24, 39, 223}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x e^{\text {sech}^{-1}(a x)} \, dx\) |
\(\Big \downarrow \) 6889 |
\(\displaystyle \frac {\int 1dx}{2 a}+\frac {\sqrt {\frac {1}{a x+1}} \sqrt {a x+1} \int \frac {1}{\sqrt {1-a x} \sqrt {a x+1}}dx}{2 a}+\frac {1}{2} x^2 e^{\text {sech}^{-1}(a x)}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\sqrt {\frac {1}{a x+1}} \sqrt {a x+1} \int \frac {1}{\sqrt {1-a x} \sqrt {a x+1}}dx}{2 a}+\frac {1}{2} x^2 e^{\text {sech}^{-1}(a x)}+\frac {x}{2 a}\) |
\(\Big \downarrow \) 39 |
\(\displaystyle \frac {\sqrt {\frac {1}{a x+1}} \sqrt {a x+1} \int \frac {1}{\sqrt {1-a^2 x^2}}dx}{2 a}+\frac {1}{2} x^2 e^{\text {sech}^{-1}(a x)}+\frac {x}{2 a}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {\sqrt {\frac {1}{a x+1}} \sqrt {a x+1} \arcsin (a x)}{2 a^2}+\frac {1}{2} x^2 e^{\text {sech}^{-1}(a x)}+\frac {x}{2 a}\) |
Input:
Int[E^ArcSech[a*x]*x,x]
Output:
x/(2*a) + (E^ArcSech[a*x]*x^2)/2 + (Sqrt[(1 + a*x)^(-1)]*Sqrt[1 + a*x]*Arc Sin[a*x])/(2*a^2)
Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[( a*c + b*d*x^2)^m, x] /; FreeQ[{a, b, c, d, m}, x] && EqQ[b*c + a*d, 0] && ( IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Int[E^ArcSech[(a_.)*(x_)^(p_.)]*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*(E^ ArcSech[a*x^p]/(m + 1)), x] + (Simp[p/(a*(m + 1)) Int[x^(m - p), x], x] + Simp[p*(Sqrt[1 + a*x^p]/(a*(m + 1)))*Sqrt[1/(1 + a*x^p)] Int[x^(m - p)/( Sqrt[1 + a*x^p]*Sqrt[1 - a*x^p]), x], x]) /; FreeQ[{a, m, p}, x] && NeQ[m, -1]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.09 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.24
method | result | size |
default | \(\frac {\sqrt {-\frac {a x -1}{a x}}\, x \sqrt {\frac {a x +1}{a x}}\, \left (x \sqrt {-a^{2} x^{2}+1}\, \operatorname {csgn}\left (a \right ) a +\arctan \left (\frac {\operatorname {csgn}\left (a \right ) a x}{\sqrt {-a^{2} x^{2}+1}}\right )\right ) \operatorname {csgn}\left (a \right )}{2 \sqrt {-a^{2} x^{2}+1}\, a}+\frac {x}{a}\) | \(92\) |
Input:
int((1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))*x,x,method=_RETURNVERBOSE)
Output:
1/2*(-(a*x-1)/a/x)^(1/2)*x*((a*x+1)/a/x)^(1/2)*(x*(-a^2*x^2+1)^(1/2)*csgn( a)*a+arctan(csgn(a)*a*x/(-a^2*x^2+1)^(1/2)))/(-a^2*x^2+1)^(1/2)*csgn(a)/a+ x/a
Time = 0.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.07 \[ \int e^{\text {sech}^{-1}(a x)} x \, dx=\frac {a^{2} x^{2} \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} + 2 \, a x - \arctan \left (\sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}}\right )}{2 \, a^{2}} \] Input:
integrate((1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))*x,x, algorithm="fricas" )
Output:
1/2*(a^2*x^2*sqrt((a*x + 1)/(a*x))*sqrt(-(a*x - 1)/(a*x)) + 2*a*x - arctan (sqrt((a*x + 1)/(a*x))*sqrt(-(a*x - 1)/(a*x))))/a^2
\[ \int e^{\text {sech}^{-1}(a x)} x \, dx=\frac {\int 1\, dx + \int a x \sqrt {-1 + \frac {1}{a x}} \sqrt {1 + \frac {1}{a x}}\, dx}{a} \] Input:
integrate((1/a/x+(-1+1/a/x)**(1/2)*(1+1/a/x)**(1/2))*x,x)
Output:
(Integral(1, x) + Integral(a*x*sqrt(-1 + 1/(a*x))*sqrt(1 + 1/(a*x)), x))/a
\[ \int e^{\text {sech}^{-1}(a x)} x \, dx=\int { x {\left (\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}\right )} \,d x } \] Input:
integrate((1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))*x,x, algorithm="maxima" )
Output:
x/a + integrate(sqrt(a*x + 1)*sqrt(-a*x + 1), x)/a
\[ \int e^{\text {sech}^{-1}(a x)} x \, dx=\int { x {\left (\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}\right )} \,d x } \] Input:
integrate((1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))*x,x, algorithm="giac")
Output:
integrate(x*(sqrt(1/(a*x) + 1)*sqrt(1/(a*x) - 1) + 1/(a*x)), x)
Time = 32.43 (sec) , antiderivative size = 303, normalized size of antiderivative = 4.09 \[ \int e^{\text {sech}^{-1}(a x)} x \, dx=\frac {\ln \left (\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2}+1\right )\,1{}\mathrm {i}}{2\,a^2}-\frac {\ln \left (\frac {\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}}{\sqrt {\frac {1}{a\,x}+1}-1}\right )\,1{}\mathrm {i}}{2\,a^2}+\frac {\frac {1{}\mathrm {i}}{32\,a^2}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2\,1{}\mathrm {i}}{16\,a^2\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2}-\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^4\,15{}\mathrm {i}}{32\,a^2\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^4}}{\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2}+\frac {2\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^4}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^4}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^6}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^6}}+\frac {x}{a}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2\,1{}\mathrm {i}}{32\,a^2\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2} \] Input:
int(x*((1/(a*x) - 1)^(1/2)*(1/(a*x) + 1)^(1/2) + 1/(a*x)),x)
Output:
(log(((1/(a*x) - 1)^(1/2) - 1i)^2/((1/(a*x) + 1)^(1/2) - 1)^2 + 1)*1i)/(2* a^2) - (log(((1/(a*x) - 1)^(1/2) - 1i)/((1/(a*x) + 1)^(1/2) - 1))*1i)/(2*a ^2) + (1i/(32*a^2) + (((1/(a*x) - 1)^(1/2) - 1i)^2*1i)/(16*a^2*((1/(a*x) + 1)^(1/2) - 1)^2) - (((1/(a*x) - 1)^(1/2) - 1i)^4*15i)/(32*a^2*((1/(a*x) + 1)^(1/2) - 1)^4))/(((1/(a*x) - 1)^(1/2) - 1i)^2/((1/(a*x) + 1)^(1/2) - 1) ^2 + (2*((1/(a*x) - 1)^(1/2) - 1i)^4)/((1/(a*x) + 1)^(1/2) - 1)^4 + ((1/(a *x) - 1)^(1/2) - 1i)^6/((1/(a*x) + 1)^(1/2) - 1)^6) + x/a + (((1/(a*x) - 1 )^(1/2) - 1i)^2*1i)/(32*a^2*((1/(a*x) + 1)^(1/2) - 1)^2)
Time = 0.15 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.55 \[ \int e^{\text {sech}^{-1}(a x)} x \, dx=\frac {-2 \mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )+\sqrt {a x +1}\, \sqrt {-a x +1}\, a x +2 a x}{2 a^{2}} \] Input:
int((1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))*x,x)
Output:
( - 2*asin(sqrt( - a*x + 1)/sqrt(2)) + sqrt(a*x + 1)*sqrt( - a*x + 1)*a*x + 2*a*x)/(2*a**2)